STRUCTURE OF In-113
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) ( July 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks , discovered by Gell-Mann and Zweig, I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ”(2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838.68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). In this photo I present the electromagnetic laws governing the nuclear structure, but a student of Einstein (Dr Th. Kalogeropoulos ) criticised my discovery of nuclear force and structure by believing that the nuclear structure is due to the invalid relativity. In fact, here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. Nuclear structure of Indium with S = +9/2 having 26 blank positions Indium (In) consists of two primordial nuclides, with the most common (~ 95.7%) nuclide (In-115) being measurably though weakly radioactive. The stable isotope In-113 is only 4.3% of naturally occurring indium. Among elements with a known stable isotope, only tellurium and rhenium similarly occur with a stable isotope in lower abundance than the long-lived radioactive isotope. STRUCTURE OF In-113 WITH S = +9/2 ' Here the square of negative spins cannot exist because the deuterons like the p37n37 and p39n39 with S =- 2 changed their spins from S = -2 to S = +2 giving S =+4 for making horizontal bonds with p38n38 and n40p40. Since the two symmetrical alpha particles give S=0 one concludes that the spin S = +9/2 of the above stable nuclide is due to the summation of the change of spins of the squares which is S = + 4 and of the total spin S = +1/2 of the 15 extra neutrons . Here we clear that in the structure of In-98 with S = 0 the p37n37 and p39n39 with S = - 2 of the square of negative spins change the spins from S = -2 to S= + 2 because they move to the square of positive spins for making horizontal bonds with the p38n38 and n40p40. Then in the presence of 8 extra neutrons of positive spins and 7 extra neutrons of negative spins one gets S = +9/2. That is S = +4 + 8(+1/2) + 7(-1/2) = +9/2. ' ''' '''DIAGRAM OF INDIUM-113 FORMING 26 BLANK POSITIONS Here you see the p47n47 along with the p48n48, which make two symmetrical alpha particles of opposite spins . But you cannot see the additional p49n49 of S=0 which makes a vertical rectangle with the n15p17. Moreover you cannot see the p37n37 and the p39n39 existing in front of p38n38 and behind the n40p40. Note that the change of their spins in In-98 with S=0 from S =-2 to S =+2 gives S =+4 Also the p41, n41, p42, n42, p43, n43 p44, and n44 which form the central parallelepiped of opposite spins are not shown. Here also the 8 deuterons of opposite spins from p13n13 to p20n20 and the 4 deuterons from p33n33 to p36 n36 are not shown. Moreover the 4 extra neutrons (n) of the first and the sixth plane are not shown, while the extra neutrons 2n existing over the p31 and p32, along with the extra 4n existing near the p23, p24, p29 and p30, are shown. ' n40.........p40......n ' ' n.........p38..........n38 H. square with 2n ' ' n31………p12.........n12.......p32' ' p31........n11.........p11…… n32 Sixth H. plane' ' n........ p29.........n10.........p10…… n30' ' n29………p9..........n9 …….p30..........n Fifth H. plane' ' p47.......n27.........p8..........n8.........p28......... n48' ' n45.........p27.........n7..........p7........n28..........p46 Fourth H. plane' ' n47.........p25.........n6.........p6..........n26...........p48' ' p45......n25……….p5..........n5……….p26........n46 Third H. plane' ' n23………p4........n4………….p24...........n' ' n........p23……..n3………p3………..n24 Second H.plane' ' p21.........n2………p2............n22' ' n21........p1........n1.........p22 First H. plane' ' ' ' ' TOP VIEW OF THE FIRST HORIZONTAL PLANE IN WHICH ALL NUCLEONS ARE SHOWN ' HERE THE FIRST EXTRA NEUTRON (n ) MAKES THE TWO RADIAL BONDS WITH p22 AND p33 WHILE THE SECOND ONE MAKES THE TWO RADIAL BONDS WITH p21 AND p34 ' ' ' (n)........p34....... n34 ' p21....... n2........ p2....... n22 ' ' n21.........p1. .......n1.......p22 ' ' n33.......p33..... (n)' Category:Fundamental physics concepts